The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment, the displacement of the wire is y1=Asin(πx/L)sinωt and energy is E1, and in another experiment its displacement is y2=Asin(2πx/L)sin2ωt and energy is E2. Then
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a
E2=E1
b
E2=2E1
c
E2=4E1
d
E2=16E1
answer is C.
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Detailed Solution
Energy (E) ∝ (Amplitude)2 (Frequency)2 Amplitude is same in both the cases, but frequency 2 ω in the second case is two times the frequency (ω) in the first case. Hence E2=4E1.